The velocity of an object with a mass of #3 kg# is given by #v(t)= - 5sin 2 t + cos 7 t #. What is the impulse applied to the object at #t= pi /6 #?

1 Answer
Feb 9, 2016

Answer:

#int F*d t=-10,098 "N.s"#

Explanation:

#v(t)=-5sin2t+cos7t#
#d v=(-10cos2t-7sin7t)d t#
#int F*d t=int m*d v#
#int F* d t=m int(-10cos2t-7sin7t)d t#
#int F* d t=m(-5sint+cos7t )#
#int F*d t=3((-5sin pi)/6+cos(7pi)/6) #
#int F*d t=3(-5*0,5-0,866)#
#int F*d t=3(-2,5-0,866)#
#int F*d t=-10,098 "N.s"#